Laws of Motion
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Question : 3
Total: 40
Give the magnitude and direction of the net force acting on a stone of mass 0.1 kg,
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km h– 1
(c) just after it is dropped from the window of a train accelerating with 1 m s– 2
(d) lying on the floor of a train which is accelerating with 1 m s– 2
Neglect air resistance throughout.
(a) just after it is dropped from the window of a stationary train,
(b) just after it is dropped from the window of a train running at a constant velocity of 36 km h
(c) just after it is dropped from the window of a train accelerating with 1 m s
(d) lying on the floor of a train which is accelerating with 1 m s
Neglect air resistance throughout.
Solution:
(a) Here, m = 0.1 kg, a = + g = 10 m s– 2
Net force, F = ma = 0.1 × 10 = 1.0 N
This force acts vertically downwards.
(b) When the train is running at a constant velocity, its acceleration = 0.
No force acts on the stone due to this motion. Therefore, force on thestone
F = weight of stone = mg = 0.1 × 10 = 1.0 N
This force also acts vertically downwards.
When the train is accelerating with 1 m s– 2
F′ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction.
But once the stone is dropped from the train, F′ becomes zero and the netforce on the stone is
F = mg = 0.1 × 10 = 1.0 N, acting vertically downwards.
(d) As the stone is lying on the floor of the train, its acceleration is sameas that of the train.
∴ Force acting on stone, F = ma = 0.1 × 1 = 0.1 N
This force is along the horizontal direction of motion of the train. Note thatweight of the stone in this case is being balanced by the normal reaction.
Net force, F = ma = 0.1 × 10 = 1.0 N
This force acts vertically downwards.
(b) When the train is running at a constant velocity, its acceleration = 0.
No force acts on the stone due to this motion. Therefore, force on thestone
F = weight of stone = mg = 0.1 × 10 = 1.0 N
This force also acts vertically downwards.
When the train is accelerating with 1 m s
F′ = ma = 0.1 × 1 = 0.1 N acts on the stone in the horizontal direction.
But once the stone is dropped from the train, F′ becomes zero and the netforce on the stone is
F = mg = 0.1 × 10 = 1.0 N, acting vertically downwards.
(d) As the stone is lying on the floor of the train, its acceleration is sameas that of the train.
∴ Force acting on stone, F = ma = 0.1 × 1 = 0.1 N
This force is along the horizontal direction of motion of the train. Note thatweight of the stone in this case is being balanced by the normal reaction.
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